# Show that if a Banach space is generated by a countable set with finite dimension then it has finite dimension

Mathematics Asked on January 5, 2022

I don’t know what to do with the following question:

Show that if a Banach space is generated by a countable set with finite dimension then it has finite dimension

Supposedly it’s related to the Baire theorem, but that didn’t help me one bit. In fact it looks totally unrelated:

Theorem (Baire) If $$(G_n)_{n geq 1}$$ is a sequence of open, dense subsets of the complete metric space $$(X,d)$$, then the intersection $$bigcap_{n geq 1} G_n$$ is dense in $$X$$.

So how would one go solving this?

However, I do know how to show what you have on your title using Baire's theorem so here's a sketch:

I'll leave to you to check all the details.

The following statement is a direct consequence of Baire's theorem

Theorem. If $$(X,d)$$ is a complete metric space and $$X=bigcup_{n geq 1} X_n$$, then $$mathrm{int}(overline{X_n}) neq varnothing$$ for some $$n$$.

Now I will use the theorem to show that the algebraic dimension of a Banach space is either finite or uncountable:

Assume that $$X$$ is a Banach space generated by $${ xi_n : n in mathbb{Z}_{>0}}$$. For each $$n in mathbb{Z}_{>0}$$ let $$X_n$$ be the space generated by $${xi_1, ldots, xi_n}$$. Now it's standard to show that

• $$X_n$$ is closed
• $$mathrm{int}(X_n)= varnothing$$

Since $$X$$ is generated by these $$X_n$$'s, we have $$X=bigcup_{n geq 1} X_n$$. Thus, by the theorem above $$mathrm{int}(X_n) neq varnothing$$ for some $$n$$, a contradiction.

Answered by Alonso Delfín on January 5, 2022

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